Problem: The perimeter of a rectangle is $22\text{ cm}$ and its area is $24\text{ cm}^2$. What is the smaller of the two integer dimensions of the rectangle?
Answer: Let $l$ and $w$ represent the length and width of our rectangle. We know that $2l+2w=22\text{ cm}$ from the information given. Dividing both sides by 2, we see that the equation simplifies to $l+w=11\text{ cm}$. We are also given that $lw=24\text{ cm}^2$. At this point, we could solve individually for $l$ or $w$, but that would quickly involve quadratics and other relatively messy things for this type of question. The two values sum to $11\text{ cm}$, so they can't be that large. Having $l$ or $w$ be equal to just $1\text{ cm}$ is much too small for the second equation. $(l,w)=(2,9)$ satisfies the first, but falls short of the second. However, if we have either $l$ or $w$ be $\boxed{3\text{ cm}}$, then both equations can be satisfied.